# If the real part of $f$ is bounded above by $A$ in the unit disk, then $|f(z)| \leq {2A|z|}/({1-|z|})$

Let $f$ be a holomorphic function on unit disk $\mathbb{D}$ such that $f(0)=0$ and there exists constant $A>0$ such that $\operatorname{Re}(f(z)) \leq A$, for all $z \in \mathbb{D}.$ Could anyone advise me how to prove $|f(z)| \leq \dfrac{2A|z|}{1-|z|}, \forall z\in \mathbb{D \ ?}$ I tried using the theorem of Schwarz pick to no avail. Hints will suffice, thank you very much.

• I see your point. – AnyAD Oct 6 '14 at 10:46

More generally: suppose $f:\mathbb D\to \Omega$ is a holomorphic map into a simply connected domain $\Omega$, and let $\phi:\mathbb D\to\Omega$ be a bijective holomorphic map such that $\phi(0)=f(0)$. By the Schwarz lemma applied to $\phi^{-1}\circ f$ we have $(\phi^{-1}\circ f)( \mathbb D_r)\subset \mathbb D_r$ for $0<r<1$. Rewrite this as $f(\mathbb D_r)\subset \phi(\mathbb D_r)$, and it begins to look promising.
To find $\phi$ for your particular domain, think of fractional linear transformations.